Task concerning Stone Weierstrass












0












$begingroup$


Let $D$ be the open unit circle with center $0$ in $mathbb{C}$.



Let $overline D$ be the closed unit circle with center $0$ in $mathbb{C}$.



Let $mathbb{C}[z]_{big|overline D}$ be the set of all functions $f:overline Dtomathbb{C}$ of polynomials of the type $f(z)=sum_{j=0}^{N}a_jz^j$.



The task has something to do with the complex version of Stone-Weierstrass.




I need to show which of the conditions for the sentence are not fulfilled:




$mathbb{C}[z]_{big|overline D}$ is a algebra in $C_b(D,mathbb{C})$ where $C_b(D,mathbb{C})$ describes the set of all continuous and bounded functions from $D$ to $mathbb{C}$. I think there should be no problem because $mathbb{C}[z]_{big|overline D}$ is a linear subspace and for $f,g in mathbb{C}[z]_{big|overline D}$ $Rightarrow fg in mathbb{C}[z]_{big|overline D}$.



Then the next one: $z_1 neq z_2 Rightarrow f(z_1) neq f(z_2)$ should be no problem either by choosing $f(z)=z$ And for every $z$ there exists a $f$ such that $f(z)neq 0$ by simply choosing $f=1$ constant.



So If I didn't make any mistakes so far the only thing that can go wrong is $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$.



$overline f(z)=sum_{j=0}^{N}overline a_j z^{-j}$. Is this not in $mathbb{C}[z]_{big|overline D}$ because $z$ can have negative exponents?






real-analysis complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 16:38










  • $begingroup$
    So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
    $endgroup$
    – conrad
    Dec 15 '18 at 16:51










  • $begingroup$
    I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
    $endgroup$
    – conrad
    Dec 15 '18 at 17:00












  • $begingroup$
    No, it isn't${}$.
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 21:20
















0












$begingroup$


Let $D$ be the open unit circle with center $0$ in $mathbb{C}$.



Let $overline D$ be the closed unit circle with center $0$ in $mathbb{C}$.



Let $mathbb{C}[z]_{big|overline D}$ be the set of all functions $f:overline Dtomathbb{C}$ of polynomials of the type $f(z)=sum_{j=0}^{N}a_jz^j$.



The task has something to do with the complex version of Stone-Weierstrass.




I need to show which of the conditions for the sentence are not fulfilled:




$mathbb{C}[z]_{big|overline D}$ is a algebra in $C_b(D,mathbb{C})$ where $C_b(D,mathbb{C})$ describes the set of all continuous and bounded functions from $D$ to $mathbb{C}$. I think there should be no problem because $mathbb{C}[z]_{big|overline D}$ is a linear subspace and for $f,g in mathbb{C}[z]_{big|overline D}$ $Rightarrow fg in mathbb{C}[z]_{big|overline D}$.



Then the next one: $z_1 neq z_2 Rightarrow f(z_1) neq f(z_2)$ should be no problem either by choosing $f(z)=z$ And for every $z$ there exists a $f$ such that $f(z)neq 0$ by simply choosing $f=1$ constant.



So If I didn't make any mistakes so far the only thing that can go wrong is $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$.



$overline f(z)=sum_{j=0}^{N}overline a_j z^{-j}$. Is this not in $mathbb{C}[z]_{big|overline D}$ because $z$ can have negative exponents?






real-analysis complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 16:38










  • $begingroup$
    So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
    $endgroup$
    – conrad
    Dec 15 '18 at 16:51










  • $begingroup$
    I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
    $endgroup$
    – conrad
    Dec 15 '18 at 17:00












  • $begingroup$
    No, it isn't${}$.
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 21:20














0












0








0





$begingroup$


Let $D$ be the open unit circle with center $0$ in $mathbb{C}$.



Let $overline D$ be the closed unit circle with center $0$ in $mathbb{C}$.



Let $mathbb{C}[z]_{big|overline D}$ be the set of all functions $f:overline Dtomathbb{C}$ of polynomials of the type $f(z)=sum_{j=0}^{N}a_jz^j$.



The task has something to do with the complex version of Stone-Weierstrass.




I need to show which of the conditions for the sentence are not fulfilled:




$mathbb{C}[z]_{big|overline D}$ is a algebra in $C_b(D,mathbb{C})$ where $C_b(D,mathbb{C})$ describes the set of all continuous and bounded functions from $D$ to $mathbb{C}$. I think there should be no problem because $mathbb{C}[z]_{big|overline D}$ is a linear subspace and for $f,g in mathbb{C}[z]_{big|overline D}$ $Rightarrow fg in mathbb{C}[z]_{big|overline D}$.



Then the next one: $z_1 neq z_2 Rightarrow f(z_1) neq f(z_2)$ should be no problem either by choosing $f(z)=z$ And for every $z$ there exists a $f$ such that $f(z)neq 0$ by simply choosing $f=1$ constant.



So If I didn't make any mistakes so far the only thing that can go wrong is $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$.



$overline f(z)=sum_{j=0}^{N}overline a_j z^{-j}$. Is this not in $mathbb{C}[z]_{big|overline D}$ because $z$ can have negative exponents?






real-analysis complex-analysis







share|cite|improve this question











$endgroup$




Let $D$ be the open unit circle with center $0$ in $mathbb{C}$.



Let $overline D$ be the closed unit circle with center $0$ in $mathbb{C}$.



Let $mathbb{C}[z]_{big|overline D}$ be the set of all functions $f:overline Dtomathbb{C}$ of polynomials of the type $f(z)=sum_{j=0}^{N}a_jz^j$.



The task has something to do with the complex version of Stone-Weierstrass.




I need to show which of the conditions for the sentence are not fulfilled:




$mathbb{C}[z]_{big|overline D}$ is a algebra in $C_b(D,mathbb{C})$ where $C_b(D,mathbb{C})$ describes the set of all continuous and bounded functions from $D$ to $mathbb{C}$. I think there should be no problem because $mathbb{C}[z]_{big|overline D}$ is a linear subspace and for $f,g in mathbb{C}[z]_{big|overline D}$ $Rightarrow fg in mathbb{C}[z]_{big|overline D}$.



Then the next one: $z_1 neq z_2 Rightarrow f(z_1) neq f(z_2)$ should be no problem either by choosing $f(z)=z$ And for every $z$ there exists a $f$ such that $f(z)neq 0$ by simply choosing $f=1$ constant.



So If I didn't make any mistakes so far the only thing that can go wrong is $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$.



$overline f(z)=sum_{j=0}^{N}overline a_j z^{-j}$. Is this not in $mathbb{C}[z]_{big|overline D}$ because $z$ can have negative exponents?






real-analysis complex-analysis




real-analysis complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 17:00







conrad

















asked Dec 15 '18 at 16:34









conradconrad

757




757












  • $begingroup$
    It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 16:38










  • $begingroup$
    So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
    $endgroup$
    – conrad
    Dec 15 '18 at 16:51










  • $begingroup$
    I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
    $endgroup$
    – conrad
    Dec 15 '18 at 17:00












  • $begingroup$
    No, it isn't${}$.
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 21:20


















  • $begingroup$
    It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 16:38










  • $begingroup$
    So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
    $endgroup$
    – conrad
    Dec 15 '18 at 16:51










  • $begingroup$
    I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
    $endgroup$
    – conrad
    Dec 15 '18 at 17:00












  • $begingroup$
    No, it isn't${}$.
    $endgroup$
    – Saucy O'Path
    Dec 15 '18 at 21:20
















$begingroup$
It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
$endgroup$
– Saucy O'Path
Dec 15 '18 at 16:38




$begingroup$
It is true that $sum_{j=0}^N overline a_j z^{-j}$ is not in $left.Bbb C[z]rightrvert_{overline D}$, but it's not true that $overline f(z)=sum_{j=0}^N overline a_j z^{-j}$ (unless $N=0$). In fact, $$overline f(z)=sum_{j=0}^N overline a_j overline z^{j}.$$
$endgroup$
– Saucy O'Path
Dec 15 '18 at 16:38












$begingroup$
So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
$endgroup$
– conrad
Dec 15 '18 at 16:51




$begingroup$
So for $f in mathbb{C}[z]_{big|overline D} Rightarrow overline f in mathbb{C}[z]_{big|overline D}$ is fulfilled?
$endgroup$
– conrad
Dec 15 '18 at 16:51












$begingroup$
I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
$endgroup$
– conrad
Dec 15 '18 at 17:00






$begingroup$
I'm wondering, is $overline f$ a polynomial of the type $f(z)=sum_{j=0}^{N}a_jz^j$?
$endgroup$
– conrad
Dec 15 '18 at 17:00














$begingroup$
No, it isn't${}$.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 21:20




$begingroup$
No, it isn't${}$.
$endgroup$
– Saucy O'Path
Dec 15 '18 at 21:20










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